aeronautical simulation..

Discussion in 'Architecture & Engineering' started by scifes, Jun 30, 2009.

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  1. Oli Heute der Enteteich... Registered Senior Member

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    The software you're looking for isn't available as a single package.

    Nope.
    You have to stick to scientific facts.
     
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  3. scifes In withdrawal. Valued Senior Member

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    i meant zero g relative to earth, meaning it will get out of earth's gravitational influence..at this time i don't care for it being in equilibrium, it may wander off in space for eternity for all i care..jest get free from earth's g.

    GREAT, this saves me a loooot of research, but wait a minute..

    what does the underlined mean?

    what if it's 1km/h?
    will it orbit the earth?

    orbiting the earth is easier than escaping the grav. pull?(i guess that's common sense; ONLY if stuff can orbit earth WHILE still in it's grav. reach.. is that possible?)

    i think the lighter something is, the smaller is it's orbiting diameter? right?




    i'm thinking just float and cut of some (hopefully most) of the distance a rocket needs to..hence less fuel requirement..

    ???
    in which direction?

    just cut the string of an elevator..

    i'm hoping two(solid works and mathworks-which i didn't find btw, looks like i'll stick to comsol or matlab-) would suffice.


    of course, i'll have to..

    but the methods are what i'm talking about.
     
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  5. Oli Heute der Enteteich... Registered Senior Member

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    MathCad, not Mathworks.
     
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  7. Sarkus Hippomonstrosesquippedalo phobe Valued Senior Member

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    Okay. So outside the earth's sphere of influence.

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    Hooooo - where to start.

    If you throw something horizontally, it falls toward earth at 'g' (9.81m/s2) (ignoring air-resistance etc).
    But the earth is curved... so if you throw it fast enough the object will fall at the same rate that the earth curves, and thus never land. This is, basically, orbiting.
    It is more complex than that, but this helped me understand it many years ago.

    Now, in order to get into low-earth orbit (c.100km above the surface) then you need to be travelling at 7km/s or so...
    If you are only travelling at 1km/h then you will never reach even a few feet off the ground, let alone an orbit.
    If you are in orbit and cut your speed to 1km/h then you will fall to earth.

    No. When considering a small object orbiting something vastly larger then the mass of the small object can be ignored - and orbital diameter is purely a function of speed and the mass of the planet. If you are in a circular orbit and you increase your speed then you will enter a larger but elliptic orbit. If you slow down you will be in a small elliptic orbit. (You would need to burn twice to change from a circular to circular orbit.)

    Yep - could work - but to launch anything significant would require the balloons to be far larger than anything ever produced before. And the energy required might not be as efficient as either multi-stage rockets or piggy-backing on a launch vehicle (e.g. Burt Rutan's sub-orbital X-prize winner).

    Also, bear in mind that the heighest a balloon has reached is around 53km (30 miles). I'm not sure how high you would actually be able to get carrying a sizeable payload.

    Circular - i.e. velocity is always tangential to the orbit.

    Seriously, you should open up a text book or even just google "orbital mechanics" or some such and get at least a basic understanding of what you're after.

    You could, but it's somewhat riskier.

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    Last edited: Jul 15, 2009
  8. PieAreSquared Woo is resistant to reason Registered Senior Member

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    Mathcad..owned by PTC now... the makers of ProE... why would you want to use SaladWorks??...ProE kicks it's ass in 3d solids
     
  9. Oli Heute der Enteteich... Registered Senior Member

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  10. PieAreSquared Woo is resistant to reason Registered Senior Member

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    money is no object Oli

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  11. mugaliens Registered Member

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    I concur. Pure, real-time, blade-analysis engineering, not tables like most flight simulators.
     
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